
# 7.7.1: The Method of Frobenius II (Exercises)


## Q7.6.1

In Exercises 7.6.1-7.6.11 find a fundamental set of Frobenius solutions. Compute the terms involving $$x^{n+r_1}$$, where $$0\le n\le N$$ ($$N$$ at least $$7$$) and $$r_1$$ is the root of the indicial equation. Optionally, write a computer program to implement the applicable recurrence formulas and take $$N>7$$.

1. $$x^2y''-x(1-x)y'+(1-x^2)y=0$$

2. $$x^2(1+x+2x^2)y'+x(3+6x+7x^2)y'+(1+6x-3x^2)y=0$$

3. $$x^2(1+2x+x^2)y''+x(1+3x+4x^2)y'-x(1-2x)y=0$$

4. $$4x^2(1+x+x^2)y''+12x^2(1+x)y'+(1+3x+3x^2)y=0$$

5. $$x^2(1+x+x^2)y''-x(1-4x-2x^2)y'+y=0$$

6. $$9x^2y''+3x(5+3x-2x^2)y'+(1+12x-14x^2)y=0$$

7. $$x^2y''+x(1+x+x^2)y'+x(2-x)y=0$$

8. $$x^2(1+2x)y''+x(5+14x+3x^2)y'+(4+18x+12x^2)y=0$$

9. $$4x^2y''+2x(4+x+x^2)y'+(1+5x+3x^2)y=0$$

10. $$16x^2y''+4x(6+x+2x^2)y'+(1+5x+18x^2)y=0$$

11. $$9x^2(1+x)y''+3x(5+11x-x^2)y'+(1+16x-7x^2)y=0$$

## Q7.6.2

In Exercises 7.6.12-7.6.22 find a fundamental set of Frobenius solutions. Give explicit formulas for the coefficients.

12. $$4x^2y''+(1+4x)y=0$$

13. $$36x^2(1-2x)y''+24x(1-9x)y'+(1-70x)y=0$$

14. $$x^2(1+x)y''-x(3-x)y'+4y=0$$

15. $$x^2(1-2x)y''-x(5-4x)y'+(9-4x)y=0$$

16. $$25x^2y''+x(15+x)y'+(1+x)y=0$$

17. $$2x^2(2+x)y''+x^2y'+(1-x)y=0$$

18. $$x^2(9+4x)y''+3xy'+(1+x)y=0$$

19. $$x^2y''-x(3-2x)y'+(4+3x)y=0$$

20. $$x^2(1-4x)y''+3x(1-6x)y'+(1-12x)y=0$$

21. $$x^2(1+2x)y''+x(3+5x)y'+(1-2x)y=0$$

22. $$2x^2(1+x)y''-x(6-x)y'+(8-x)y=0$$

## Q7.6.3

In Exercises 7.6.23-7.6.27 find a fundamental set of Frobenius solutions. Compare the terms involving $$x^{n+r_{1}}$$, where $$0\leq n\leq N$$ ($$N$$ at least $$7$$) and $$r_{1}$$ is the root of the indicial equation. Optionally, write a computer program to implement the applicable recurrence formulas and take $$N>7$$.

23. $$x^2(1+2x)y''+x(5+9x)y'+(4+3x)y=0$$

24. $$x^2(1-2x)y''-x(5+4x)y'+(9+4x)y=0$$

25. $$x^2(1+4x)y''-x(1-4x)y'+(1+x)y=0$$

26. $$x^2(1+x)y''+x(1+2x)y'+xy=0$$

27. $$x^2(1-x)y''+x(7+x)y'+(9-x)y=0$$

## Q7.6.4

In Exercises 7.6.28-7.6.38 find a fundamental set of Frobenius solutions. Give explicit formulas for the coefficients.

28. $$x^2y''-x(1-x^2)y'+(1+x^2)y=0$$

29. $$x^2(1+x^2)y''-3x(1-x^2)y'+4y=0$$

30. $$4x^2y''+2x^3y'+(1+3x^2)y=0$$

31. $$x^2(1+x^2)y''-x(1-2x^2)y'+y=0$$

32. $$2x^2(2+x^2)y''+7x^3y'+(1+3x^2)y=0$$

33. $$x^2(1+x^2)y''-x(1-4x^2)y'+(1+2x^2)y=0$$

34. $$4x^2(4+x^2)y''+3x(8+3x^2)y'+(1-9x^2)y=0$$

35. $$3x^2(3+x^2)y''+x(3+11x^2)y'+(1+5x^2)y=0$$

36. $$4x^2(1+4x^2)y''+32x^3y'+y=0$$

37. $$9x^2y''-3x(7-2x^2)y'+(25+2x^2)y=0$$

38. $$x^2(1+2x^2)y''+x(3+7x^2)y'+(1-3x^2)y=0$$

## Q7.6.5

In Exercises 7.6.39-7.6.43 find a fundamental set of Frobenius solutions. Compute the terms involving $$x^{2m+r_{1}}$$, where $$0 ≤ m ≤ M$$ ($$M$$ at least $$3$$) and $$r_{1}$$ is the root of the indicial equation. Optionally, write a computer program to implement the applicable recurrence formulas and take $$M > 3$$.

39. $$x^2(1+x^2)y''+x(3+8x^2)y'+(1+12x^2)y$$

40. $$x^2y''-x(1-x^2)y'+(1+x^2)y=0$$

41. $$x^2(1-2x^2)y''+x(5-9x^2)y'+(4-3x^2)y=0$$

42. $$x^2(2+x^2)y''+x(14-x^2)y'+2(9+x^2)y=0$$

43. $$x^2(1+x^2)y''+x(3+7x^2)y'+(1+8x^2)y=0$$

## Q7.6.6

In Exercises 7.6.44-7.6.52 find a fundamental set of Frobenius solutions. Give explicit formulas for the coefficients.

44. $$x^2(1-2x)y''+3xy'+(1+4x)y=0$$

45. $$x(1+x)y''+(1-x)y'+y=0$$

46. $$x^2(1-x)y''+x(3-2x)y'+(1+2x)y=0$$

47. $$4x^2(1+x)y''-4x^2y'+(1-5x)y=0$$

48. $$x^2(1-x)y''-x(3-5x)y'+(4-5x)y=0$$

49. $$x^2(1+x^2)y''-x(1+9x^2)y'+(1+25x^2)y=0$$

50. $$9x^2y''+3x(1-x^2)y'+(1+7x^2)y=0$$

51. $$x(1+x^2)y''+(1-x^2)y'-8xy=0$$

52. $$4x^2y''+2x(4-x^2)y'+(1+7x^2)y=0$$

## Q7.6.7

53. Under the assumptions of Theorem 7.6.2, suppose the power series

$\sum_{n=0}^\infty a_n(r_1)x^n \quad\mbox{ and }\quad \sum_{n=1}^\infty a_n'(r_1)x^n\nonumber$

converge on $$(-\rho,\rho)$$.

1. Show that $y_1=x^{r_1}\sum_{n=0}^\infty a_n(r_1)x^n\quad\mbox{ and }\quad y_2=y_1\ln x+x^{r_1}\sum_{n=1}^\infty a_n'(r_1)x^n\nonumber$ are linearly independent on $$(0,\rho)$$. HINT: Show that if $$c_{1}$$ and $$c_{2}$$ are constants such that $$c_{1}y_{1}+c_{2}y_{2}≡0$$ on $$(0, \rho )$$, then $(c_{1}+c_{2}\ln x)\sum_{n=0}^{\infty}a_{n}(r_{1})x^{n}+c_{2}\sum_{n=1}^{\infty}a_{n}'(r_{1})x^{n}=0,\quad 0<x<\rho \nonumber$ Then let $$x\to 0+$$ to conclude that $$c_{2}=0$$.
2. Use the result of (a) to complete the proof of Theorem 7.6.2.

54. Let

$Ly=x^2(\alpha_0+\alpha_1x)y''+x(\beta_0+\beta_1x)y'+(\gamma_0+\gamma_1x)y\nonumber$

and define

$p_0(r)=\alpha_0r(r-1)+\beta_0r+\gamma_0\quad\mbox{ and }\quad p_1(r)=\alpha_1r(r-1)+\beta_1r+\gamma_1.\nonumber$

Theorem 7.6.1 and Exercise 7.5.55a

imply that if

$y(x,r)=x^r\sum_{n=0}^\infty a_n(r)x^n\nonumber$

where

$a_n(r)=(-1)^n\prod_{j=1}^n{p_1(j+r-1)\over p_0(j+r)},\nonumber$

then

$Ly(x,r)=p_0(r)x^r.\nonumber$

Now suppose $$p_0(r)=\alpha_0(r-r_1)^2$$ and $$p_1(k+r_1)\ne0$$ if $$k$$ is a nonnegative integer.

1. Show that $$Ly=0$$ has the solution $y_1=x^{r_1}\sum_{n=0}^\infty a_n(r_1)x^n,\nonumber$ where $a_n(r_1)={(-1)^n\over\alpha_0^n(n!)^2}\prod_{j=1}^np_1(j+r_1-1).\nonumber$
2. Show that $$Ly=0$$ has the second solution $y_2=y_1\ln x+x^{r_1}\sum_{n=1}^\infty a_n(r_1)J_nx^n,\nonumber$ where $J_n=\sum_{j=1}^n{p_1'(j+r_1-1)\over p_1(j+r_1-1)}-2\sum_{j=1}^n{1\over j}.\nonumber$
3. Conclude from (a) and (b) that if $$\gamma_1\ne0$$ then $y_1=x^{r_1}\sum_{n=0}^\infty {(-1)^n\over(n!)^2}\left(\gamma_1\over\alpha_0\right)^nx^n\nonumber$ and $y_2=y_1\ln x-2x^{r_1}\sum_{n=1}^\infty {(-1)^n\over(n!)^2}\left(\gamma_1\over\alpha_0\right)^n \left(\sum_{j=1}^n{1\over j}\right)x^n\nonumber$ are solutions of $\alpha_0x^2y''+\beta_0xy'+(\gamma_0+\gamma_1x)y=0.\nonumber$ (The conclusion is also valid if $$\gamma_1=0$$. Why?)

55. Let

$Ly=x^2(\alpha_0+\alpha_qx^q)y''+x(\beta_0+\beta_qx^q)y'+(\gamma_0+\gamma_qx^q)y\nonumber$

where $$q$$ is a positive integer, and define

$p_0(r)=\alpha_0r(r-1)+\beta_0r+\gamma_0\quad\mbox{ and }\quad p_q(r)=\alpha_qr(r-1)+\beta_qr+\gamma_q.\nonumber$

Suppose

$p_0(r)=\alpha_0(r-r_1)^2 \quad\mbox{ and }\quad p_q(r)\not\equiv0.\nonumber$

1. Recall from Exercise 7.5.59 that $$Ly~=0$$ has the solution $y_1=x^{r_1}\sum_{m=0}^\infty a_{qm}(r_1)x^{qm},\nonumber$ where $a_{qm}(r_1)={(-1)^m\over (q^2\alpha_0)^m(m!)^2}\prod_{j=1}^mp_q\left(q(j-1)+r_1\right).\nonumber$
2. Show that $$Ly=0$$ has the second solution $y_2=y_1\ln x+x^{r_1}\sum_{m=1}^\infty a_{qm}'(r_1)J_mx^{qm},\nonumber$ where $J_m=\sum_{j=1}^m{p_q'\left(q(j-1)+r_1\right)\over p_q\left(q(j-1)+r_1\right)}-{2\over q}\sum_{j=1}^m{1\over j}.\nonumber$
3. Conclude from (a) and (b) that if $$\gamma_q\ne0$$ then $y_1=x^{r_1}\sum_{m=0}^\infty {(-1)^m\over(m!)^2}\left(\gamma_q\over q^2\alpha_0\right)^mx^{qm}\nonumber$ and $y_2=y_1\ln x-{2\over q}x^{r_1}\sum_{m=1}^\infty {(-1)^m\over(m!)^2}\left(\gamma_q\over q^2\alpha_0\right)^m\left(\sum_{j=1}^m{1\over j}\right)x^{qm}\nonumber$ are solutions of $\alpha_0x^2y''+\beta_0xy'+(\gamma_0+\gamma_qx^q)y=0.\nonumber$

56. The equation

$xy''+y'+xy=0\nonumber$

is Bessel’s equation of order zero. (See Exercise 7.5.53.) Find two linearly independent Frobenius solutions of this equation.

57. Suppose the assumptions of Exercise 7.5.53 hold, except that

$p_0(r)=\alpha_0(r-r_1)^2.\nonumber$

Show that

$y_1={x^{r_1}\over\alpha_0+\alpha_1x+\alpha_2x^2}\quad\mbox{ and }\quad y_2={x^{r_1}\ln x\over\alpha_0+\alpha_1x+\alpha_2x^2}\nonumber$

are linearly independent Frobenius solutions of

$x^2(\alpha_0+\alpha_1x+\alpha_2 x^2)y''+x(\beta_0+\beta_1x+\beta_2x^2)y'+ (\gamma_0+\gamma_1x+\gamma_2x^2)y=0\nonumber$

on any interval $$(0,\rho)$$ on which $$\alpha_0+\alpha_1x+\alpha_2x^2$$ has no zeros.

## Q7.6.8

58. $$4x^2(1+x)y''+8x^2y'+(1+x)y=0$$

59. $$9x^2(3+x)y''+3x(3+7x)y'+(3+4x)y=0$$

60. $$x^2(2-x^2)y''-x(2+3x^2)y'+(2-x^2)y=0$$

61. $$16x^2(1+x^2)y''+8x(1+9x^2)y'+(1+49x^2)y=0$$

62. $$x^2(4+3x)y''-x(4-3x)y'+4y=0$$

63. $$4x^2(1+3x+x^2)y''+8x^2(3+2x)y'+(1+3x+9x^2)y=0$$

64. $$x^2(1-x)^2y''-x(1+2x-3x^2)y'+(1+x^2)y=0$$

65. $$9x^2(1+x+x^2)y''+3x(1+7x+13x^2)y'+(1+4x+25x^2)y=0$$

## Q7.6.9

66.

1. Let $$L$$ and $$y(x,r)$$ be as in Exercises 7.5.57 and 7.5.58. Extend Theorem 7.6.1 by showing that $L\left({\partial y\over \partial r}(x,r)\right)=p'_0(r)x^r+x^rp_0(r)\ln x.\nonumber$
2. Show that if $p_0(r)=\alpha_0(r-r_1)^2\nonumber$ then $y_1=y(x,r_1) \quad \text{and} \quad y_2={\partial y\over\partial r}(x,r_1)\nonumber$ are solutions of $$Ly=0$$.